Question 146600: Soybean meal is 18% protein; corn meal is 9% protein. How many pounds of each should be mixed together inorder to get 360-lb mixture that is 14% protein? How many pounds of cornmeal should be in the mixture?
How many pounds of soybean should be in the mixture?
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Let's start by assigning variables to represent the unknown quantities.
Let S = the number of pounds of Soybean meal required.
Let C = the number of pounds of Corn meal required.
The amount of protein in S pounds of Soybean mean can be represented as: 18% of S or, (0.18)S
The amount of protein in the C pounds of Corn meal can be represented as: 9% of C or, 0.09C
Now if we add these two amounts of protein, we should get 14% of 360 or (0.14)360
Now we can write the equation:
(0.18)S+(0.09)C = (0.14)360
But we have two unknowns and only one equation, so, to remove one of the unknowns (S or C, let's choose C), we can see that the number of pounds of C is just 360 - S, so we'll substitute C = 360-S and rewrite the equation.
(0.18)S + (0.09)(360-S) = (0.14)360 Now we can solve for C.
(0.18)S + 32.4 - (0.09)S = 50.4 Combine like-terms on the left side.
(0.09)S +32.4 = 50.4 Subtract 32.4 from both sides.
(0.09)S = 18 Finally, divide both sides by 0.09
S = 200 and
C = 360-S = 360-200 = 160
The mixture of 360 pounds of 14% protein should contain 200 pounds of Soybean meal and 160 pounds of Corn meal.
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